Visual Plane
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What survives

Two triangles and the line nobody drew

Desargues' theorem is the reason a hand-drawn shadow construction closes. An object and its shadow are two figures in perspective from the lamp; the theorem says their corresponding sides meet, pairwise, on one line — which is the ground line. So the closure a draughtsman treats as confirmation that the work is accurate is a theorem they cannot violate.
17 min read 11 figures Projected, not constructedOne machine used twiceWhat a projection destroys

Worth reading first: Where parallel lines meet · The diagonals find the middle.

Two triangles are in perspective from a point when the three lines joining their corresponding vertices all pass through one place. That is a strong condition and it is easy to arrange: put a light at that point and let one triangle be the shadow of the other.

Desargues, in 1639, noticed that such a pair satisfies a second condition nobody put there. Pair off the corresponding sides and intersect each pair. The three points that come out lie on one line.

Two triangles in perspective from a pointCorresponding vertices lie on three lines through one centre. Pair off the corresponding SIDES instead and the three points where they meet are collinear — 3e-13 px from the line through them, at every configuration the slider reaches. Nothing was measured to make that happen, and nothing can be adjusted to improve it.three side intersections, collinear to 3e-13 pxcentrethree sides paired, three pointscollinear to 3e-13 px
Fig. 1 Two triangles in perspective from a point. Corresponding vertices lie on three lines through one centre; corresponding sides meet at three points, and those three are collinear to arithmetic noise — at every configuration the slider reaches, with nothing measured to make it happen.

There is no measurement anywhere in the statement. Both halves are about incidence: which lines pass through which points. The theorem asserts that one incidence pattern forces another, and the forcing is exact.

Checked, and shown to discriminate

Forty configurations, the centre and the three vertices and the three scale factors all moving, and the largest distance from any of the three intersections to the line through them stays at the level of arithmetic noise throughout.

That is worth having and it is only half a test. An assertion that has never rejected anything means nothing, so the machinery is also fed two triangles that are not in perspective from a point, and their three side-intersections miss collinearity by 5.08 px — a distance no drawing could hide. The theorem discriminates: it is a property of triangles in perspective and not a property of any three intersections whatsoever.

Why a shadow construction closes

Here is the reason this belongs on a site about pictures rather than in a geometry course.

Take a triangular plate standing on the floor and a lamp above it. The plate and its shadow are two triangles, and they are in perspective from the lamp by construction — each vertex, its shadow, and the lamp are collinear, because that is what casting a shadow is.

So Desargues applies. Extend an edge of the plate and the corresponding edge of the shadow until they meet; do it for all three pairs; the three meeting points are collinear. And the line they are on is not arbitrary — it is the line where the plate’s own plane cuts the floor.

A box and its shadow, both projections from a pointThe rays from the lamp to the corners are the same construction as the rays from the eye to the corners — one operation, two centres.correct from 26 cm, at 160 mm wide34° across
Fig. 2 The construction the theorem is about. A solid and its shadow, both computed from one centre of projection, the lamp. Every edge and its shadow meet on the ground line where the object’s plane crosses the floor, and the meeting is forced rather than achieved.

This is the fact hand construction relies on without naming. A draughtsman building a shadow by hand runs an edge out to the ground line, runs the shadow’s edge out to meet it, and takes the meeting as confirmation that the construction is accurate. It is not confirmation of anything. The three points would be collinear whatever numbers went in, as long as the construction really was made from one centre — so closure tests that the draughtsman used one lamp, and nothing else.

That is the same category of trap as the cross-ratio test that four equally spaced posts also pass, and it is worth stating in the same words: a necessary condition that a wrong answer satisfies is not a check. Closure is necessary. It is satisfied by every construction from a single centre, including the ones with the lamp in the wrong place.

Two versions of the same invariant, one of which measures nothingFour consecutive divisions give the equal-steps method a perfect score. Using the vanishing point as the fourth point rejects it by 14%.error against the value the projection must producefour divisionsthree plus the VPthe projectionexactexactequal stepsexact14%halving3.6%25%tapering0.9%21%green: agrees with the projectiona necessary condition is not a test
Fig. 3 The same shape of trap, measured in the wrong field. A test that a wrong answer passes is not a test — and the working version has to be built so that the wrong answer fails it.
The outline, the shadow, and the outline againA point light is a centre of projection, so the map from the occluder's plane to the floor is a plane projectivity and has an inverse. Running each shadow point back through the lamp returns the outline to 4e-16 m — the recovered curve is drawn over the original and cannot be seen apart from it.the outline, its shadow, and the outline recovered from itcorrect from 20 cm, at 160 mm widerecovered to 4e-16 m
Fig. 4 The shadow map the theorem is about, drawn as the projectivity it is. An outline, its shadow, and the outline recovered — and every incidence in the original survives into the shadow, which is what Desargues is a consequence of.

The converse, and why it matters more

Desargues’ theorem has a converse and it is the same statement read backwards: if the three side-intersections of two triangles are collinear, the two triangles are in perspective from a point.

That direction is the useful one for reading a picture. Given a photograph with an object and its cast shadow, extend three pairs of corresponding edges. If the three meeting points are collinear, the picture is consistent with a single point light; if they are not, it is not, and no amount of adjusting the light’s position will fix it.

So the theorem provides a test of a picture that needs no camera calibration, no horizon, no known lengths — only a straightedge and three pairs of edges. Which is a rare thing: most consistency tests on a photograph need something supplied.

Six posts in sunlight from 40°The shadows are parallel in the world, so in the picture they meet at one point on the horizon — found from the drawn shadows to 1e-12 px.horizonshadows meet at x = -198, off the frameon the horizon, as it must be
Fig. 5 The same closure under the sun. With the light at infinity the three side-intersections still lie on a line, and the centre of perspective is a vanishing point rather than a lamp — a case the theorem needs no special provision for.

The self-dual shape of it

Desargues’ theorem has a property worth pointing at even though this site does nothing with it. Its converse is its own dual: exchange the words point and line throughout, replace lies on with passes through, and the statement turns into its own converse.

Duality is not a trick of notation. In the projective plane, points and lines are genuinely interchangeable — every theorem has a partner obtained by swapping them — and it is why four concurrent lines have a cross-ratio exactly as four collinear points do. Desargues is the standard example because its dual is its converse, so the theorem and its converse are one fact.

Four points on a line, before and after a projectionLength and the ratio of lengths do not survive the projection; the cross-ratio does, agreeing to 0e+0 relative. Joined to a vertex off their line, the four points become four lines whose own cross-ratio is the same number — and two further transversals cut those lines in four points that carry it again, which is why any picture of the four rays gives the same answer.horizonABCDany vertexon the groundin the picturelength AB1.00011.3930ratio AB:CD0.56670.6837cross-ratio1.31681.3168correct from 26 cm, at 160 mm wide34° across
Fig. 6 Duality doing something. The same four points, joined to a vertex off their line, become four lines with a cross-ratio of their own — and any transversal cuts those four lines in four points carrying the same number again.
The three vanishing points of one box, drawn to scale with the boxThe picture is the small rectangle. Two of the three vanishing points fall well outside it, which is why they are computed rather than located by eye.the pictureVP₁VP₂VP₃orthocentrefocal length from the triangle — 739.9 pxspread 3e-14% across three routes
Fig. 7 Three pencils in one picture. Each is a set of drawn lines whose common point is a direction rather than a place, and every incidence theorem on this site depends on those points being ordinary points of the geometry.

The configuration, counted

The theorem’s picture has a tidy census worth doing, because the tidiness is the reason the statement is self-dual rather than merely happening to have a dual.

Count the points: three vertices of the first triangle, three of the second, the centre of perspective, and the three intersections of corresponding sides. Ten points. Count the lines: three sides of each triangle, three lines through the centre joining corresponding vertices, and the axis the three intersections lie on. Ten lines.

Now count incidences. Every one of the ten points lies on exactly three of the ten lines, and every one of the ten lines passes through exactly three of the ten points. A configuration with those numbers maps to itself when points and lines are exchanged, which is exactly what being self-dual means — and it is why dualising Desargues produces its own converse rather than a different theorem.

The census also says why no vertex is privileged. The centre of perspective looks special in the drawing because it is where the construction starts, and in the configuration it is one of ten points on three lines each, indistinguishable from the three intersections that are supposed to be the conclusion. Choose a different point of the ten to call the centre and the same ten points and lines are a Desargues configuration again, with two different triangles.

What the theorem does not say

Three things it is regularly taken to say and does not, and each of them is a way a picture can be inconsistent while passing the test.

It does not say the two triangles are similar, or related by any measurable transformation. They are related by a projection from a point, which can take any triangle to any other; the theorem constrains only the incidences. A shadow that is the right shape and the wrong size passes.

It does not say where on the axis the three points fall. Only that they are on one line. Two of the three could be a hand’s breadth apart and the third a mile away, and the picture is still consistent with a single centre. The metric content is nil.

And it does not say the centre is where it looks. The construction locates the centre as the common point of three lines, and three lines through one point in the picture are three lines through one point in space only if the picture is a projection at all. A drawing made by hand can have three concurrent lines and no consistent lamp behind them, because concurrency is one condition and a lamp is three numbers.

That last one is the reason this rung sits under the lamp recovery rather than beside it. Desargues says the shadow construction closes; the recovery says where the lamp is and whether every post in the picture agrees about it. The first is necessary and the second is the measurement.

The theorem’s other instances on this site

A shadow is the obvious instance and it is not the only one. Three more appear in fields that have nothing to do with light, and listing them is the fastest way to see what the theorem is actually about.

A reflection. A scene and its mirror image are in perspective from — strictly, are related by a reflection through — the mirror plane, and the drawn figure and its drawn reflection are two triangles whose corresponding vertices lie on lines perpendicular to the mirror. Their corresponding sides meet on the mirror’s own line, which is why a reflection construction closes for the same reason a shadow construction does.

A photograph of a photograph. The original print and the rephotograph are related by a homography, and a homography takes triangles to triangles preserving all incidences, so any Desargues configuration in one is a Desargues configuration in the other.

And a scaled drawing. Two concentric similar triangles are in perspective from their common centre, and the theorem says their corresponding sides meet on a line — which for a pure scaling is the line at infinity, since corresponding sides are parallel. That is the degenerate case, and it is a case rather than an exception only because points at infinity are ordinary points.

The common element is not light and not photography. It is a map that preserves incidence, which is what a projection from a centre is — and the theorem is a fact about such maps rather than about any of the things that happen to be one.

What it means for a picture that fails

The converse gives a test, and a test is only useful if a failure says something. It says three things, and they can be told apart.

A small failure — a few pixels of non-collinearity — is measurement noise. Extending an edge to find an intersection amplifies any error in locating the edge, and the amplification grows as the two edges become more nearly parallel. A picture whose three points miss a line by a distance comparable to the width of a drawn line has failed nothing.

A large failure with the three points scattered is more than one light source. Each pair of corresponding sides is consistent with some centre, and if the centres differ the three intersections have no reason to line up.

And a large failure with the three points on a smooth curve rather than a line is a curved surface. The shadow of a plane figure onto a curved receiver is not a projectivity, so its edges are not straight and the intersections are not where a straight-edge construction puts them. That is the same failure the un-casting rung measures with a four-point homography, arriving through the incidences instead of through the map.

Three distinguishable failure modes from one straightedge construction is a good return, and none of them needs a camera, a length or a horizon. The test is available on any photograph containing an object and its cast shadow, which is most photographs taken outdoors.

Why the theorem is harder than it looks

There is a fact about Desargues’ theorem that reads as a curiosity and is a real statement about what “projective plane” means.

In three dimensions the theorem is nearly trivial. Two triangles in perspective from a point, lying in two different planes, have corresponding sides that meet — because each pair of corresponding sides lies in a common plane, the one through the centre and the pair — and every intersection lies in both triangles’ planes, which meet in a line. Three points on the line where two planes meet. Done in a sentence.

In two dimensions it is not trivial at all, and it is not even always true: there are projective planes, perfectly consistent as axiom systems, in which Desargues’ theorem fails. What makes it true in the plane of an ordinary drawing is precisely that the plane can be embedded in a three-dimensional space — that the drawing is a picture of something.

So the theorem holds here for the reason this site would want it to hold. Every picture on this site is a projection of a three-dimensional scene through a centre, and the flat statement is the three-dimensional one seen from that centre. A drawing plane that could not be embedded that way would be a plane no camera could photograph, and Desargues is one of the things that would go wrong.

Where it fails, and what that says

The theorem is true in the projective plane. In the drawn plane it can appear to fail, and the way it fails is instructive.

Two corresponding sides can be parallel in the picture. Then they do not meet, their intersection is a point at infinity, and there is nothing to draw. The theorem is untouched — the point at infinity is on the line just like the others, and the line through the remaining two intersections is exactly the line whose direction that parallel pair defines — but a construction carried out on paper stops working, and a program that intersects the pairs has to decide what to return.

This machinery returns a finite point when there is one and refuses when the pair is parallel. Refusing is right: a very distant intersection is a real answer and a parallel pair has none, and the difference between the two is the difference between a number and a direction.

That is the same distinction the whole vanishing ladder is built on. Where parallel lines meet is about promoting the direction to a point so that it can enter calculations like any other; here the promotion is exactly what makes Desargues unconditional. Without points at infinity the theorem needs three special cases; with them it needs none.

A family of parallel ground lines at 22°, and where they meetAll five lines pass through one point on the horizon, off the edge of the frame at x = 1188. The point fitted from the drawn lines agrees with the one computed from the direction to 3e-12 px, and the fit's own residual is 6e-13 px.horizon — the image of the line at infinityvanishing point at x = 1188 — off the framecorrect from 26 cm, at 160 mm wide34° across
Fig. 8 The promotion the theorem depends on. A direction becomes an ordinary point of the picture, findable from the drawn lines and enterable into any calculation the others are — including the calculation of whether three points are collinear.
Halving a receding rectangle two waysThe diagonals cross at the image of the rectangle's centre, 6e-14 px from it — the construction is exact at every camera because it uses only which lines meet where, and that is what a projection keeps. Halving the drawn side with a ruler instead lands 9.4 px from the image of the side's midpoint. The same diagonal continued lays out 2 more bays of the same 2.1 m, with nothing measured.the diagonals against a ruler, at 5.0 mthe diagonals — exactthe ruler — 9.4 px outcorrect from 23 cm, at 160 mm wideharmonic set -1.000000000 · construction 6e-14 px
Fig. 9 The first member of the family, continued outward. Incidence alone, exact at every camera, and no measurement anywhere in it.

The pattern this rung belongs to

Three constructions on this site are exact because they are made of incidence alone, and it is worth listing them together because the shared property is more useful than any of them individually.

The diagonals of a rectangle cross at its centre, so the diagonals of the image cross at the image of the centre — an exact halving with no measurement.

Desargues, so a shadow construction closes and a picture’s lighting can be tested with a straightedge.

And the pole of the horizon is the image of a circle’s centre, so a photograph of a round table carries the position of its centre exactly.

None of these needs a camera, a length, a horizon in units, or a calibration. All three survive photography because they never asked for anything a photograph destroys. That is what “projective” means in practice, and it is a much more useful characterisation than any definition of the projective plane: a projective statement is one that survives somebody photographing the drawing and throwing the original away.

Two triangles in perspective from a pointCorresponding vertices lie on three lines through one centre. Pair off the corresponding SIDES instead and the three points where they meet are collinear — 2e-13 px from the line through them, at every configuration the slider reaches. Nothing was measured to make that happen, and nothing can be adjusted to improve it.three side intersections, collinear to 2e-13 pxcentrethree sides paired, three pointscollinear to 2e-13 px
Fig. 10 The same theorem with the axis not drawn, at a different configuration. The three points are still collinear; the line is a conclusion rather than a construction line, which is why the figure can leave it out and lose nothing.
Halving a receding rectangle two waysThe diagonals cross at the image of the rectangle's centre, 6e-14 px from it — the construction is exact at every camera because it uses only which lines meet where, and that is what a projection keeps. Halving the drawn side with a ruler instead lands 13.6 px from the image of the side's midpoint.the diagonals against a ruler, at 4.2 mthe diagonals — exactthe ruler — 13.6 px outcorrect from 23 cm, at 160 mm wideharmonic set -1.000000000 · construction 6e-14 px
Fig. 11 The companion construction, from the rung below. Both are incidence-only, both are exact at every camera, and both do work that a measurement on the page cannot do at all.

Where this goes next

the diagonals against a ruler, at 3.2 mthe diagonals — exactthe ruler — 22.7 px outcorrect from 23 cm, at 160 mm wideharmonic set -1.000000000 · construction 8e-14 px The diagonals find the middle Three taught methods for spacing a receding row are all wrong, and the best of them misplaces a post by three and a half metres. There is a fourth, it needs no measurement and no vanishing point, and it is exact at every camera and every depth — because 'the diagonals of a rectangle cross at its centre' is a statement about which lines meet where, and that is the one kind of statement a projection cannot damage. correct from 26 cm, at 160 mm wide34° across A shadow is a second projection The construction that puts a shadow on the ground is the construction that puts the scene on the picture plane, with the lamp where the eye was. Shadow drawing is taught as a separate set of recipes and it is one operation with the centre moved, which is why the same code draws both. the outline, its shadow, and the outline recovered from itcorrect from 20 cm, at 160 mm widerecovered to 4e-16 m A shadow can be un-cast A shadow looks like a lossy record — a flattened smear with the shape half thrown away. It is nothing of the kind. The map from the occluder's plane to the floor is a plane projectivity, so it has an inverse, and the outline comes back out of its own shadow exactly. What breaks it is not the light and not the shape: it is the floor not being flat.

What links here

Computed from the collection, not written here: the essays that point at this one.

Reads more easily once this is understood

Essays that name this one as worth reading first.

Shares its objects with

Essays that name at least two of the same things, and that neither author linked.

Named objects

A flat tag is an object no other essay names yet.

centre of projectionDemonstrationDesarguesGround lineHarmonic conjugateline at infinitynecessary, not sufficientPoint lightProjective dualityProjective invariantProjective map